Derivative of Inverse Trigonometric Functions
f(x) = cos(arcsin(2x))
What is f'(x)?
I did this down below, after the one drwls answered
copied:
It is so similar that you should be able to do it the same way. Draw the triangle and you have sqrt(1-4x^2)
f(x) = (1-4x^2)^.5
f'(x) = .5 [(1-4x^2)^-.5] (-8x)
= -4x/sqrt(1-4x^2)
Oh.
But the text book I'm using gives me the answer:
Sqrt(1-4x^2)
No denominators or anything.
That is just the first step.
To find the derivative of the function f(x) = cos(arcsin(2x)), we can use the chain rule.
First, let's find the derivative of the inner function arcsin(2x). The derivative of arcsin(x) is 1/sqrt(1-x^2), so:
f'(x) = [-sin(arcsin(2x))] * [d/dx(2x)]
= -sin(arcsin(2x)) * 2
Next, we need to find the derivative of the outer function cos(x). The derivative of cos(x) is -sin(x), so:
f'(x) = -2 * sin(arcsin(2x))
Now, we have the derivative of f(x) = cos(arcsin(2x)) as:
f'(x) = -2 * sin(arcsin(2x))
Note that this expression can be simplified further, but this is the general form of the derivative.